摘要：24． 如图.现有两块全等的直角三角形纸板Ⅰ.Ⅱ.它们两直角边的长分别为1和2．将它们分别放置于平面直角坐标系中的.处.直角边在轴上．一直尺从上方紧靠两纸板放置.让纸板Ⅰ沿直尺边缘平行移动．当纸板Ⅰ移动至处时.设与分别交于点.与轴分别交于点． (1)求直线所对应的函数关系式, (2)当点是线段上的动点时.试探究: ①点到轴的距离与线段的长是否总相等?请说明理由, ②两块纸板重叠部分的面积是否存在最大值?若存在.求出这个最大值及取最大值时点的坐标,若不存在.请说明理由． 24．解:(1)由直角三角形纸板的两直角边的长为1和2. 知两点的坐标分别为． 设直线所对应的函数关系式为．······················································ 2分 有解得 所以.直线所对应的函数关系式为．··············································· 4分 (2)①点到轴距离与线段的长总相等． 因为点的坐标为. 所以.直线所对应的函数关系式为． 又因为点在直线上. 所以可设点的坐标为． 过点作轴的垂线.设垂足为点.则有． 因为点在直线上.所以有．················· 6分 因为纸板为平行移动.故有.即． 又.所以． 法一:故. 从而有． 得.． 所以． 又有．·················································· 8分 所以.得.而. 从而总有．······························································································· 10分 法二:故.可得． 故． 所以． 故点坐标为． 设直线所对应的函数关系式为. 则有解得 所以.直线所对的函数关系式为．········································ 8分 将点的坐标代入.可得．解得． 而.从而总有．··················································· 10分 ②由①知.点的坐标为.点的坐标为． ．······························································· 12分 当时.有最大值.最大值为． 取最大值时点的坐标为．··································································· 14分

网址：http://m.1010jiajiao.com/timu3_id_480161[举报]